# Making a N.C. reed switch functio as a N.O. switch

#### iONic

Joined Nov 16, 2007
1,654
Was wireing some LED lighting under my sink cabinets and seem to have bought the wrong reed relays. They are CLOSED(current flowing) when they are in proximity to one another(when cabinet door is closed, and OPEN when the cabinet door is open...............resulting in the LED's being ON when the cabinet door is closed and off when the door is open.

No, they are not configurable for N.C. or N.O.

If there any other way to get the functionality that I need from these switcher.....additional parts??

Just thought I'd ask before I send them back. It was hard to find them at a reasonable price, and these did fit the bill with this respect.

Joined Jul 18, 2013
23,644
A small relay would do it.
What current do they switch as there are relays as small in a 8pin DIP package.
Probabally easier to source N.O. reed SW's.
Max.

#### AlbertHall

Joined Jun 4, 2014
11,506
What is the power for the LEDs?

Joined Jan 15, 2015
6,066
If there any other way to get the functionality that I need from these switcher.....additional parts??
Yes but I doubt the ends would justify the means in that you would be adding more circuitry. I see Max posted as well as Albert and I would agree the best solution would be to just get new switches suitable for the load.

Ron

#### Alec_t

Joined Sep 17, 2013
12,215
You might be able to use an auxiliary magnet to bias the switch the other way.

#### AlbertHall

Joined Jun 4, 2014
11,506
You might be able to use an auxiliary magnet to bias the switch the other way.
Ah, Baldrick has a cunning plan?

#### ebp

Joined Feb 8, 2018
2,332
A reed switch that is normally closed very likely has a biasing magnet "attached." If the magnet can be removed, it will become normally open, however everything is likely potted and inaccessible.

EDIT - if it is magnetically biased to be normally closed, it may be possible to make it normally open by reversing the field direction of the external magnet - which again may be pretty much impossible in any practical way

Joined Jul 18, 2013
23,644
I have used the Hamlin versions that are made for this application, the 59135 series, they have N.O/N.C. versions.
Max.

Joined Jan 15, 2015
6,066

ak

#### iONic

Joined Nov 16, 2007
1,654
You want N/C so the proximity of a magnet opens the switch. The magnet proximity is present when the cabinet is closed so the N/C switch is open but when the cabinet is opened, magnetic field removed, the switch will close. Think about it. As drawn the switches have no magnetic field present but when the cabinet is closed the magnet is beside the switch and the switches will close, not what you want.

Something I forgot to mention is you may find reed switches at any home improvement store like Lowes or Home Depot in the US.

Now what bothers me is how do I know that when I close my refrigerator the light really does go out?

Ron
We may have to disagree on this point.
Where I think we deviate is that I am considering the "switch" is both the reed switch and magnet.

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#### iONic

Joined Nov 16, 2007
1,654
From your schematic it looks like 40 mA per switch. One 2N7000 MOSFET and one 10K resistor per switch will invert (door open = on) the power drive to the LEDs, about $1 for all 4 circuits. ak No...$0 as I have both parts.

#### AnalogKid

Joined Aug 1, 2013
9,362

Joined Jan 15, 2015
6,066
We may have to disagree on this point.
Where I think we deviate is that I am considering the "switch" is both the reed switch and magnet.
No problem as long as you know what you want it to do. I see the state as the normal state of the switch less it being activated by the magnet. Long as it does what you want life is good.

Ron

#### iONic

Joined Nov 16, 2007
1,654
You could use a small power mosfet. If you use an N channel device connect the source to the supply negative. Connect the load (LED string.) between the drain and positive supply. connect a high value resistor (100K for example.) between the gate and positive supply. Connect the NO reed switch between the gate and source. When the switch was open the gate would be biased to the positive rail cauing the mosfet to conduct. When the reed switch was closed it would short the gate to the source which would stop the mosfet from conducting. These would be a small current flowing through the resistor when the reed switch was closed. With a 12 volt supply and a 100K resistor the current would be 120 uA.

Les.
@AnalogKid @LesJones

So, let me get this straight.
Provided I rewired the circuit right in the schematic bellow.
When the cabinet door is closed, the reed switch is closed, causing a short from Gate-Source, which in turn blocks current flow through the LED's & Drain-Source. When I open the cabinet door, the reed switch opens and removes the short between G-S, allowing the MOSFET to be turned on and allowing the LED's to turn on via current through D-S.

Really just trying to figure out the simplest place to cut wires to add these components without making a web of wires. I also need to calculate the KWH of power consumption of this active MOSFET to determine if this method is a game-changer!!!!!

#### LesJones

Joined Jan 8, 2017
3,498
You say in post #10 that the LEDs are rated at 20 mA but you also say that they have a built in 130 ohm resistor. Assuming that these are white LEDs their forward voltage will be about 3.3 volts which means there will be 8.7 volts across the 130 ohm resistors so he current through the LEDs will be 8.7/130 = 67 mA which is more than 3 times their current rating
If you use 100K resistors to the mosfet gates while the doors are closed (LEDs off.) it will consume 0.038 KWH per year. ( It would be 0.38 KWH per year if you use 10K resistors.) Both of those values would probably be less than the loss in the 12 volt power supply.

Les.

#### AnalogKid

Joined Aug 1, 2013
9,362
Your understanding of the circuit is correct.

The value of R1 is a tradeoff between static, off-state current consumption, and noise immunity. The smaller the R, the better the immunity and the higher the power dissipation. For me, I'd use something in the 1.5K - 10K range. 1.5K or more keeps the power dissipation below 0.1 W, a good value for long term reliability of a 1.4-W resistor; and 10K or less reduces the circuit impedance for improved noise immunity. If the 12 Vdc source is AC line powered (as opposed to batteries), I'd go with 2.2 K.

ak